Problem: $f(x,y) = x^3y + y^2$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A, Incorrect) Incorrect $3x^2 + 2y$ (Choice B, Incorrect) Incorrect $3x^2$ (Choice C, Incorrect) Incorrect $x^3 + 2y$ (Choice D, Checked, Correct) Correct (selected) $3x^2y$
Answer: We want to find $\dfrac{\partial f}{\partial x}$, which is the partial derivative of $f$ with respect to $x$. When we take a partial derivative with respect to $x$, we treat $y$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial x} \left[ x^3y \right] = 3x^2y \\ \\ &\dfrac{\partial}{\partial x} \left[ y^2 \right] = 0 \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial x} = 3x^2y + 0 = 3x^2y$